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I'm reading Sawyer's Prelude to Mathematics, from page 34-35:

[...]Prove that, if¹

$$\frac{ac-b^2}{a-2b+c}=\frac{bd-c^2}{b-2c+c^2}$$

Then the fractions just given are both equal to:

$$\frac{ad-bc}{a-b-c+d}$$

This question has a very definite form, and obviously to hammer it our by a lengthy and sharpless calculation, while veryfing the result, would bring one no nearer to the heart of the question. What interested me most was the question was the question, how did the examiner come to think of this question?

The pattern of the question includes the following aspects, $ac-b^2=0$ is the condition for the three quantities $a$, $b$, $c$ to be in geometrical progression.

1 - It is assumed that $b$ and $c$ are unequal. The text does not discuss this point, as it's not relevant to the main theme. What suggested the question to the examiner.

I've asked something similar before and someone said me it's the discriminant of the equation but here I'm in doubt on what he meant with conditon for $a$, $b$, $c$ to be in geometrical progression.

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1  
The expression $ac-b^2$ is not a condition. The equation $ac-b^2=0$ is a condition for $a$, $b$, $c$ to be in GP. –  André Nicolas Sep 18 '12 at 5:06
    
@AndréNicolas Sorry, mistyped. I guess you kinda solved my problem. I didn't see the damn $=0$. –  Vladimir Putin Sep 18 '12 at 5:09
    
$0$ is easy to miss, it is so small. –  André Nicolas Sep 18 '12 at 5:13
    
It actually does not exist. –  Vladimir Putin Sep 18 '12 at 5:15

1 Answer 1

up vote 3 down vote accepted

We say that $a,b,c$ are in geometric progression if there is some $r$ such that $b=ra$ and $c=r^2a$, so the sequence reads $a,ra,r^2a$.

Note that unless $a=b=c=0$, $r$ must be $b/a$ so the condition for $a,b,c$ to be in geometric progression is $c=(b/a)^2a=b^2/a$, or equivalently that $b^2-ac=0$.

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I can't understand where's the $ r$ coming from. –  Vladimir Putin Sep 18 '12 at 15:17
    
@GustavoBandeira I'm not sure what you mean. The existence of some such $r$ is the definition of being in geometric progression. –  Alex Becker Sep 18 '12 at 19:58
    
I was thinking that I could switch any of the $a$, $b$, $c$ and it would always equals zero. Then I thought that this would be a little dumb, I guess it's impossible for it to hold true for all numbers. Your answer has some kind of heuristic that is still unknown to me. –  Vladimir Putin Sep 18 '12 at 20:14
    
@GustavoBandeira You mean if $b^2-ac=0$ then $a^2-bc=0$? Certainly not, consider $a=1,b=2,c=4$. –  Alex Becker Sep 18 '12 at 20:16
    
I dunno how you proceed from the condition to the thing you made with the $r$. But it's starting to make sense. –  Vladimir Putin Sep 18 '12 at 20:18

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